Proof of Three-signed Equivalence to Complex Numbers

Products in three-signed math exactly match products in complex arithmetic. 
Sums are also equivalent.

By choosing the star pole to be on the positive real axis the product arithmetic
works out. The transform from a three-signed value to a complex value is:

z = s(y) - (1/2)(m(y) + p(y)) + i(sqrt(3)/2)(m(y) - p(y))

Where y is in P3 and m(),p(),s() return the minus,plus and star
components of their argument.

These equations come from pure graphical analysis of the three-signed
branches being at angles 2pi/3. The components resolve via right
triangles. The minus pole is oriented toward the positive imaginary axis.

- minus pole
-
- +Imaginary
- |
- |
- |
- . | . . . . .p1 = * 7 - 4
- | .
- | .
. -| .
. 0 * * * * * * * * * * star pole, +Real
. + origin .
p3 = + 4 - 2 . + .
. + .
+ .
+ .
+ . . . . . . . . . p2 = * 9 + 6
+
+
+
+ plus pole




It will be shown that

z( y1 y2 ) = z( y1 ) z( y2 )

where the product on the left is performed in P3 and the product on the right is performed in C.
To simplify notation I will simply use s,m,and p to represent the
three sign components. I will also use q3 to represent the square root of
three.

We have
z1 = s1-(1/2)(m1+p1) + i(q3/2)(m1-p1)
and
z2 = s2-(1/2)(m2-p2) + i(q3/2)(m2-p2) .

The complex product z1z2 is then:

(s1-m1/2-p1/2 + iq3m1/2-iq3p1/2)(s2-m2/2-p2/2 + iq3m2/2-iq3p2/2)

= s1s2 -s1m2/2 -s1p2/2 +s1m2iq3/2-s1p2iq3/2
+ -m1s2/2 +m1m2/4 +m1p2/4 -m1m2iq3/4 +m1p2iq3/4
+ -p1s2/2 +p1m2/4 +p1p2/4 -p1m2iq3/4 +p1p2iq3/4
+ +m1s2iq3/2 -m1m2iq3/4 -m1p2iq3/4 -m1m2(3/4) +m1p2(3/4)
+ -p1s2iq3/2 +p1m2iq3/4 +p1p2iq3/4 +p1m2(3/4) -p1p2(3/4) .

Gathering terms with precedence in the order of s,m,p:

= s1s2 -s1m2/2 +s1m2iq3/2 -s1p2/2 -s1p2iq3/2
+ -m1s2/2 +m1s2iq3/2 +m1m2/4 -m1m2(3/4) -m1m2iq3/4 -m1m2iq3/4 +m1p2/4
+m1p2iq3/4 -m1p2iq3/4 +m1p2(3/4)
+ -p1s2/2 -p1s2iq3/2 +p1m2/4 +p1m2(3/4) +p1m2iq3/4 -p1m2iq3/4 +p1p2/4
-p1p2(3/4) +p1p2iq3/4 +p1p2iq3/4

and reducing cancellable terms:

= s1s2 -s1m2/2 +s1m2iq3/2 -s1p2/2 -s1p2iq3/2
+ -m1s2/2 +m1s2iq3/2 -m1m2/2 -m1m2iq3/2 +m1p2
+ -p1s2/2 -p1s2iq3/2 +p1m2 -p1p2/2 +p1p2iq3/2

[ Equation 1 ] .

The product y1y2 has the following properties based on its definition:

m = m1s2 + p1p2 + s1m2.
p = m1m2 + p1s2 + s1p2.
s = m1p2 + p1m2 + s1s2.

Converted to a complex value it is:

+m1p2 +p1m2 +s1s2 -(1/2){m1s2 +p1p2 +s1m2 +m1m2 +p1s2 +s1p2 }
+i(q3/2){m1s2 +p1p2 +s1m2 -(m1m2 + p1s2 + s1p2)}

= m1p2 +p1m2 +s1s2 -m1s2/2 -p1p2/2 -s1m2/2 -m1m2/2 -p1s2/2 -s1p2/2
+m1s2iq3/2 +p1p2iq3/2 +s1m2iq3/2 -m1m2iq3/2 -p1s2iq3/2 - s1p2iq3/2 .

Now gathering terms from the above to better match the complex product
of Equation 1:

= +s1s2 -s1m2/2 +s1m2iq3/2 -s1p2/2 - s1p2iq3/2
-m1s2/2 +m1s2iq3/2 -m1m2/2 -m1m2iq3/2 +m1p2
-p1s2/2 -p1s2iq3/2 +p1m2 -p1p2/2 +p1p2iq3/2 .

This exactly matches the reduced form of the complex product and
represents the equation of the product of two three-signed values y1y2
(where s1,m1,p1 represents y1 and s2,m2,p2 represents y2) converted to
the traditional complex representation.

Sums are also equivalent.
The proof is trivial and is already apparent from the graphical behavior.

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